# Show that $\frac{D\tilde{X}}{dt}=\sum_{i=1}^n (B_i(t)-\langle B(t),c(t)\rangle c_i(t))\frac{\partial}{\partial x_i}$.

Let $X=\sum_{i=1}^nA_i\frac{\partial}{\partial x_i}$ be a vector field in $\mathbb{R^n}$ which has a restriction $\tilde{X}$ to the sphere $S^{n-1}$. Let $c:I\rightarrow S^{n-1}$ ,$c(t)=(c_1(t),...,c_n(t))$ be a smooth curve. Define $B_i(t)=\frac{d(A_i(c(t))}{dt}$, $i=1,2,...,n$. Show that $\frac{D\tilde{X}}{dt}=\sum_{i=1}^n (B_i(t)-\langle B(t),c(t)\rangle c_i(t))\frac{\partial}{\partial x_i}$.

So the formula is $\frac{D\tilde{X}}{dt}=\sum_j\frac{dA_j(c(t))}{dt}\frac{\partial}{\partial x_i}+\sum_{i,j}\frac{dc_i(t)}{dt}A_j(c(t)\nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial x_j}$. So now I am really confused with the terms $\nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial x_j}$. What are they? How do we define them?

Hint :

(1) Note that if $X=\sum_{i=1}^n A_i\frac{\partial }{\partial x_i}$, then $\frac{\partial }{\partial x_i}$ is not coordinate vector field on sphere

(2) If $N\subset M$ and $N$ has induced metric from $M$, then covariant derivative on $N$ is a projection of convariant derivative on $M$

If $M=\mathbb{R}^n$, then covariant derivative on $M$ is a just differentiation Hence we can complete the proof

• I am still confused. How do we differentiate a vector field? Oct 18, 2016 at 2:54
• $\frac{d}{dt} X(c(t)) =\sum_i \frac{d}{dt} A_i(c(t)) \frac{\partial}{\partial x_i}$ Oct 18, 2016 at 2:56
• Ok so now since $X$ has a restriction $\sum_{i=1}^n A_i(c(t))c_i(t)=0$. I wanted to differentiate this to get the other part of the required formula. Is $\sum_{i=1}^n\frac{dA_i(c(t))}{dt}c_i(t)=0$ the differentiation of it? Oct 18, 2016 at 3:04
• $\sum_i A_i c_i=0$ so that $\ast:=\sum_i \frac{d}{dt} A_i\ c_i=-\sum_i A_i \ c_i'$ Here note that $\ast$ is normal component of $X'$ wrt tangent space of sphere Oct 18, 2016 at 3:07
• But sorry still I don't see a way to get the expression. Oct 18, 2016 at 3:18